Symmetric Matrices Yielding Unique Solution in Linear System

To determine the number of matrices $A \in \mathcal{A}$ for which the system of linear equations

$$A \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} 1 \\ 0 \\ 0 \end{bmatrix}$$

has a unique solution, we need to find the number of matrices $A \in \mathcal{A}$ such that $\det(A) \neq 0$.

A general $3 \times 3$ symmetric matrix has the form:

$$A = \begin{bmatrix} a & b & c \\ b & d & e \\ c & e & f \end{bmatrix}$$

where $a, b, c, d, e, f \in \{0, 1\}$.

The sum of all the entries of $A$ is:
$\text{Sum} = (a + d + f) + 2(b + c + e)$
We are given that five entries of $A$ are $1$ and four entries are $0$. Therefore, the sum of all entries is $5$:
$(a + d + f) + 2(b + c + e) = 5$

Since $2(b + c + e)$ is even, the sum of the diagonal entries $a + d + f$ must be an odd integer. Since $a, d, f \in \{0, 1\}$, the possible values for $a + d + f$ are $1$ or $3$.

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Case 1: $a + d + f = 3$

In this case, all diagonal entries are $1$:
$a = d = f = 1$
Then:
$2(b + c + e) = 5 - 3 = 2 \implies b + c + e = 1$
Thus, exactly one of $b, c, e$ is $1$, and the remaining two are $0$.
The number of such matrices is $\binom{3}{1} = 3$.

Let us evaluate the determinant of each:

1. If $b = 1, c = 0, e = 0$:

$$A = \begin{bmatrix} 1 & 1 & 0 \\ 1 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} \implies \det(A) = 0 \quad (\text{Rows 1 and 2 are identical})$$

1. If $c = 1, b = 0, e = 0$:

$$A = \begin{bmatrix} 1 & 0 & 1 \\ 0 & 1 & 0 \\ 1 & 0 & 1 \end{bmatrix} \implies \det(A) = 0 \quad (\text{Rows 1 and 3 are identical})$$

1. If $e = 1, b = 0, c = 0$:

$$A = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 1 \\ 0 & 1 & 1 \end{bmatrix} \implies \det(A) = 0 \quad (\text{Rows 2 and 3 are identical})$$

Hence, none of the matrices in this case have $\det(A) \neq 0$.

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Case 2: $a + d + f = 1$

In this case, exactly one diagonal entry is $1$ and the other two are $0$.
Then:
$2(b + c + e) = 5 - 1 = 4 \implies b + c + e = 2$
Thus, exactly two of $b, c, e$ are $1$ and one is $0$.

The general determinant of a symmetric matrix is:
$\det(A) = a(df - e^2) - b(bf - ce) + c(be - cd)$

There are 3 choices for which diagonal entry is $1$:

1. Subcase 2.1: $a = 1, d = 0, f = 0$
   $\det(A) = 1(0 - e^2) - b(-ce) + c(be) = -e^2 + 2bce$
   Since $b + c + e = 2$:

   • If $e = 0$, then $b = 1, c = 1$:
     $\det(A) = 0$
   • If $e = 1$, then either $(b, c) = (1, 0)$ or $(b, c) = (0, 1)$:
     $\det(A) = -(1)^2 + 2(0)(1) = -1 \neq 0$
     Thus, there are $2$ matrices with $\det(A) \neq 0$.

2. Subcase 2.2: $d = 1, a = 0, f = 0$
   By symmetry, with the diagonal $1$ at the second row:

   • If the opposite off-diagonal entry $c = 0$, then $b = 1, e = 1 \implies \det(A) = 0$.
   • If $c = 1$, then either $(b, e) = (1, 0)$ or $(0, 1) \implies \det(A) = -1 \neq 0$.
     Thus, there are $2$ matrices with $\det(A) \neq 0$.

3. Subcase 2.3: $f = 1, a = 0, d = 0$
   Similarly, with the diagonal $1$ at the third row:

   • If the opposite off-diagonal entry $b = 0$, then $c = 1, e = 1 \implies \det(A) = 0$.
   • If $b = 1$, then either $(c, e) = (1, 0)$ or $(0, 1) \implies \det(A) = -1 \neq 0$.
     Thus, there are $2$ matrices with $\det(A) \neq 0$.

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Total Count

The total number of matrices $A \in \mathcal{A}$ having a non-zero determinant is:
$2 + 2 + 2 = 6$

Since $6$ is at least $4$ but less than $7$, the correct option is (B).